Examples
(Application of Fick’s first law to steady state problems with constant diffusion
coefficient (problem 1 and part of problem 3), and Fick’s second law to nonsteady
problems with thinfilm and constant surface concentration boundary conditions, and
constant diffusion coefficient (problem 2 and part of problem 3))
Problem 1
A cubic steel tank of volume 1 liter and wall thickness 0.1 cm is used to store hydrogen with an
initial pressure of 8.7 atm.
The tank is placed in outer space at 673 K.
The hydrogen
concentration
on the steel surface is given by
cp
H
=
3
(ppm)
where c is in ppm and the hydrogen pressure, p
H
, is in atm.
The density of steel is 7.73 g/cm
3
.
The diffusion coefficient of hydrogen in steel is taken to be 10
8
cm
2
/s.
What is the rate of
pressure drop (atm/s) as a result of diffusion of hydrogen through the wall?
Solution:
Assuming the process is diffusioncontrolled, the outer surface concentration of the tank is fixed
at zero at all times.
The inner surface concentration is a function of hydrogen pressure inside the
tank.
The concentration in terms of g/cm
3
is then given by
p
p
HH
H
=×
×
=
×
32
3
2
(ppm) = 3 10
7.73
(g/cm
10
(g/cm
6
3
5
3
).
)
Since the process is very slow and low, we can approximate this problem using the steady state
Fick’s first law.
JD
dc
dx
p
p
H
H
=
×
×



10
0
2 32 10
01
23 10
8
5
12
.
.
.
The total flux of hydrogen diffusing out from the tank is given by
JJ
A
p
p
p
tot
H
H
H
==×
×
(29
=
×


2 3 10
6 1000
1 38 10
6 9 10
12
1 3
2
91
0
..
.
(g/s)
mol/s
From mass balance, the total flux out of the tank is the equal to the change in the amount of
hydrogen per unit time within the tank, i.e.,
J
dn
dt
tot
=
where n is the number of moles of hydrogen.
Assuming hydrogen gas behave ideally, we have
J
dn
dt
V
RT
dp
dt
tot
H
==
where V is the volume of the tank, R is gas constant and T is temperature.
Therefore,
dp
dt
RT
V
Jp
p
H
tot
H
H
×
××
0 082
673
1
69 10
38 10
10
8
.
.
.
(atm/s)
It can be easily seen that the pressure dropping rate is proportional to the square root of the
hydrogen pressure inside the tank.
The initial pressure is 8.7 atm, so the initial rate of pressure
decrease will be
dp
dt
p
H
×
=
×
38
87 112 10
87
.
.
.
.
(atm/s)

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View Full DocumentProblem 2
There are two ways of introducing dopants into silicon by diffusion.
The first is called
predeposition during which the surface concentration of dopant atoms is maintained constant by
a vapor source; and the second is called the redistribution or drivein which is used to move
predeposited dopants to the desired junction depth.
a) Assume the substrate is ntype with a background doping level of 1.5x10
16
atoms/cm
3
and the
diffusion coefficient of boron in silicon is given by
D
B
(
T
)
=
0.76exp

3.46
eV
k
B
T
[
cm
2
/
s
]
where k
B
is the Boltzmann constant and T is temperature.
The boron surface concentration is
maintained at a constant value of 1.8x10
20
atoms/cm
3
during predeposition at 950˚C in a neutral
ambient.
If predeposition is performed for 30 min., calculate: (I) the diffusion profile; (II) the
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 Spring '02
 CHEN
 Chemistry, Sulfur, Boron, ZnS

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